# Poisson's equation and Green's functions

1. Nov 26, 2007

### jmb

Trying to verify the general integral solution to Poisson's equation I reach the following contradiction, what am I missing/doing wrong?

A solution $$u(\mathbf{x})$$ to Poisson's equation satisfies:

$$\nabla^2 u(\mathbf{x}) = - \frac{\rho(\mathbf{x})}{\epsilon_0}$$

we can find such a solution in a given domain by evaluating:

$$u(\mathbf{x}) = \int_{V} \frac{\rho(\mathbf{x'})}{4 \pi \epsilon_0 |\mathbf{x}-\mathbf{x'}|} d\mathbf{x'}$$

or equivalently (if we are only interested in the electric field):

$$\mathbf{E} = -\nabla u = - \int_{V} \frac{\rho(\mathbf{x'})}{4 \pi \epsilon_0} \frac{\mathbf{x}-\mathbf{x'}}{|\mathbf{x}-\mathbf{x'}|^3} d\mathbf{x'}$$

We can think of these solutions as either the Green's function solution or, physically, as the convolution of the point source solution with the source density $$\rho(\mathbf{x})$$.

However if I take the Laplacian (or the divergence in the case of the electric field solution) of either of the above I get zero instead of $$-\frac{\rho}{\epsilon_0}$$. What am I missing??

To evaluate the Laplacian/divergence of the above I make use of the fact that $$\mathbf{x'}$$ is a 'dummy variable', which allows me to switch the order of integration and differentiation when applying $$\nabla^2$$ to the integral on the RHS and also means the components of $$\mathbf{x'}$$ all have derivative zero. With these assumptions the operation is fairly straightforward to carry out, but I have anyway verified my answer with Mathematica.

2. Nov 26, 2007

### marcusl

Not so. You should take the Laplacian with respect to x (after all you are calculating $$\nabla^2{u(x)}$$). Since
$$\nabla^2 \frac{1}{|x-x'|}=\delta(x-x')$$
you get the right result from the integration.
This expression is derived in many places:
Jackson, Classical Electrodynamics, 2nd ed., Eq. (1.31)
Arfken, Math Methods for Physicists, 2nd ed., sect. 1.15
to give two examples.

Last edited: Nov 26, 2007
3. Nov 26, 2007

### jmb

Thanks marcusl.

Yes I was taking the derivative wrt x (hence the derivatives of all components of x' being zero). But I was erroneously evaluating the Laplacian of $$\frac{1}{|x-x'|}$$ as zero.

I've since found references to the result you quote, but they all start by searching for the Green's function $$G$$ that satisfies $$\nabla^2 G = \delta(x-x')$$ and then deducing it to be the above... out of interest do you know of any way to show it starting from the other end --- i.e. directly differentiating $$\frac{1}{|x-x'|}$$ and showing its Laplacian to be the delta function? Or do the references you quote do it that way?

I'll look them up when I get the chance. Thanks again.

4. Nov 26, 2007

### marcusl

Yes, Jackson shows it explicitly. And, now that I'm looking at his formula instead of relying on memory, I see I forgot the constant
$$\nabla^2 \frac{1}{|x-x'|}=-4\pi \delta(x-x')$$

5. Nov 26, 2007

### samalkhaiat

Last edited: Nov 26, 2007
6. Nov 27, 2007

### jmb

OK samalkhaiat I'll have a shot at your "remaining 8 marks", let me know if I'm missing something!

From the divergence theorem we have,

$$I = \int_{V} \nabla^2 (1/r) dV = \int_{S} \nabla (1/r) \cdot \mathbf{dS}$$

Since the RHS is a surface integral we don't need to worry about evaluating $$\nabla (1/r)$$ at the origin and so, choosing V as a sphere centred on the origin, we can straightforwardly write the surface integral as:

$$I = \int_{0}^{2\pi} \int_{0}^{\pi} \frac{\partial}{\partial r} \left(\frac{1}{r}\right) r^2 \sin{\theta} \; d\theta \; d\phi = -4\pi$$

(since $$\frac{\partial}{\partial r} \left(\frac{1}{r}\right) = -\frac{1}{r^2}$$) as required.

7. Nov 27, 2007

### marcusl

Jackson has similar expressions on p. 39 (he calls the integral C instead of I).

8. Nov 27, 2007

### samalkhaiat

9. Nov 28, 2007

### jmb

10. Nov 29, 2007

### samalkhaiat

Last edited: Nov 29, 2007
11. Nov 30, 2007

### jmb

Cool. Many thanks again to both marcusl and samalkhaiat for your help!